What is operator T& in std::reference_wrapper<T> ?

Suri

5y725

What is operator T& in std::reference_wrapper ??
I can't figure it out, how to apply this operator (I don't want to use get()).
source: https://en.cppreference.com/w/cpp/utility/functional/reference_wrapper/get

2 Replies

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Rajanikant Hawaldar5y ago
Accepted answer
It is the same as get. It can be used to cast a reference_wrapper to T& or when you use a reference_wrapper where a T& is expected. It is just a conversion operator.
Sachin Singh5y ago
std::reference_wrapper
is an object that imitates a reference (T&),
template<typename N>
void change(N n) {
//if n is std::reference_wrapper,
// it implicitly converts to int& here.
n += 1;
}

void foo() {
int x = 10;

int& xref = x;
change(xref); //Passed by value
//x is still 10
std::cout << x << "\n"; //10

//Be explicit to pass by reference
change<int&>(x);
//x is 11 now
std::cout << x << "\n"; //11

//Or use std::ref
change(std::ref(x)); //Passed by reference
//x is 12 now
std::cout << x << "\n"; //12
}